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Find $\mathrm{Gal}(\mathbb{F}_{p^n}/\mathbb{F}_{p^m})$, where $m|n$.

Is this proof correct?

Since the base field is finite, $\mathbb{F}_{p^m}\subset \mathbb{F}_{p^n}$ is a Galois extension. (Is there another easy way of seeing why $\mathbb{F}_{p^n}/\mathbb{F}_{p^m}$ is a Galois extension?) Therefore $|\mathrm{Gal}(\mathbb{F}_{p^n}/\mathbb{F}_{p^m})| = [\mathbb{F}_{p^n}:\mathbb{F}_{p^m}]$. But $[\mathbb{F}_{p^n}:\mathbb{F}_{p}]= [\mathbb{F}_{p^n}:\mathbb{F}_{p^m}][\mathbb{F}_{p^n}:\mathbb{F}_{p}]$. Given the claim that $[\mathbb{F}_{p^k}:\mathbb{F}_{p}] = k$ for all $k$ we finally arrive at the answer $[\mathbb{F}_{p^n}:\mathbb{F}_{p^m}]=d$. The Galois group is cyclic (since the base field was finite) so $\mathrm{Gal}(\mathbb{F}_{p^n}/\mathbb{F}_{p^m}) = \mathbb{Z}_d$

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So, you already know that the Galois group of a finite extension of a finite field must be cyclic? – Arturo Magidin Aug 26 '11 at 18:10

Modulo knowing that the Galois group of a finite extension of a finite field is cyclic, the argument is correct, though you can do much better than just use $d$: you know that $[\mathbb{F}_{p^n}:\mathbb{F}_{p^m}] = \frac{n}{m}$ (you never say what $d$ is).

Also, you may not want to say that $[\mathbb{F}_{p^k}\colon \mathbb{F}_p]=k$ is a "claim"; that implies that this is something we have not established and are only using as an undischarged assumption. Better to call it a "result" if you have already proven it.

To answer your questions: Is there an easy way to see that $\mathbb{F}_{p^n}/\mathbb{F}_{p^m}$ is a Galois extension? $\mathbb{F}_{p^n}$ is the splitting field of $x^{p^n}-x$ over $\mathbb{F}_p$; this is a separable polynomial, so $\mathbb{F}_{p^n}$ is Galois over $\mathbb{F}_p$, and hence also over any intermediate field.

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